$$\int \sqrt{\dfrac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$$
My attempt is as follows:-
$$\int \sqrt{\dfrac{(1-\sin x)(2-\sin x)(1-\sin x)}{(1+\sin x)(2+\sin x)(1-\sin x)}}dx$$
$$\int \dfrac{1-\sin x}{\cos x}\cdot\sqrt{\dfrac{(2-\sin x)}{(2+\sin x)}}dx$$ $$\int \dfrac{1-\sin x}{\cos x}\cdot\dfrac{(2-\sin x)}{\sqrt{4-\sin^2x}}dx$$
$$\int \dfrac{\cos x}{1+\sin x}\cdot\dfrac{(2-\sin x)}{\sqrt{4-\sin^2x}}dx$$
$$\sin x=t$$
$$\int \dfrac{2-t}{(1+t)\sqrt{4-t^2}}dt$$ $$\int \dfrac{3}{(1+t)\sqrt{4-t^2}}dt-\int\dfrac{dt}{\sqrt{4-t^2}}$$
$$3\int\left(\dfrac{1+t-t}{(1+t)\sqrt{4-t^2}}\right)dt-\sin^{-1}\dfrac{t}{2}$$
$$3\int \dfrac{1}{\sqrt{4-t^2}}dt+3\int\dfrac{-t}{(1+t)\sqrt{4-t^2}}dt-\sin^{-1}\dfrac{t}{2}$$
$$\sqrt{4-t^2}=y$$ $$\dfrac{-2t}{2\sqrt{4-t^2}}=\dfrac{dy}{dt}$$
$$2\sin^{-1}\dfrac{t}{2}+3\int\dfrac{1}{1+\sqrt{4-y^2}}dy$$
$$y=2\sin\theta$$ $$\dfrac{dy}{d\theta}=2\cos\theta$$
$$2\sin^{-1}\dfrac{t}{2}+3\int\dfrac{2\cos\theta}{1+2\cos\theta}d\theta$$
$$2\sin^{-1}\dfrac{t}{2}+3\theta+3\int\dfrac{1}{1+2\cos\theta}d\theta$$
$$2\sin^{-1}\dfrac{t}{2}+3\theta+3\int\dfrac{\sec^2\theta}{1+\tan^2\theta+2(1-\tan^2\theta)}d\theta$$
$$2\sin^{-1}\dfrac{t}{2}+3\theta+3\int\dfrac{\sec^2\theta}{3-\tan^2\theta}d\theta$$
$$2\sin^{-1}\dfrac{t}{2}+3\theta+3\int\dfrac{dz}{z^2-3}$$
$$2\sin^{-1}\dfrac{t}{2}+3\theta+\dfrac{\sqrt{3}}{2}\ln\left|\dfrac{z-\sqrt{3}}{z+\sqrt{3}}\right|+C$$
$$2\sin^{-1}\left(\dfrac{\sin x}{2}\right)+3\sin^{-1}\dfrac{\sqrt{4-\sin^2x}}{2}+\dfrac{\sqrt{3}}{2}\ln\left|\dfrac{y-\sqrt{3}\sqrt{4-y^2}}{y+\sqrt{3}\sqrt{4-y^2}}\right|+C$$
$$2\sin^{-1}\left(\dfrac{\sin x}{2}\right)+3\sin^{-1}\dfrac{\sqrt{4-\sin^2x}}{2}+\dfrac{\sqrt{3}}{2}\ln\left|\dfrac{\sqrt{4-\sin^2x}-\sqrt{3}\sin x}{\sqrt{4-\sin^2x}+\sqrt{3}\sin x}\right|+C$$
This solution got very long, is there any nice trick or some good solution I am missing here?
